Hess’s law• Hess’s Law states that the heat of a whole

reaction is equivalent to the sum of it’s steps.• For example: C + O2 → CO2 (pg. 165)The book tells us that this can occur as 2 steps

C + ½O2 → CO ΔH° = – 110.5 kJCO + ½O2 → CO2 ΔH° = – 283.0 kJ

C + CO + O2 → CO + CO2 ΔH° = – 393.5 kJI.e. C + O2 → CO2 ΔH° = – 393.5 kJ• Hess’s law allows us to add equations.• We add all reactants, products, & ΔH° values.• We can also show how these steps add

together via an “enthalpy diagram” …

Steps in drawing enthalpy diagrams1. Balance the equation(s).2. Sketch a rough draft based on ΔH° values.3. Draw the overall chemical reaction as an

enthalpy diagram (with the reactants on one line, and the products on the other line).

4. Draw a reaction representing the intermediate step by placing the relevant reactants on a line.

5. Check arrows: Start: two leading awayFinish: two pointing to finishIntermediate: one to, one away

6. Look at equations to help complete balancing (all levels must have the same # of all atoms).

7. Add axes and ΔH° values.

C + O2 → CO2 ΔH° = – 393.5 kJ

Reactants

Intermediate

Products

C + O2

CO2

CO

Ent

halp

y

Note: states such as (s) and (g) have been ignored to reduce clutter on these slides. You should include these in your work.

ΔH° = – 110.5 kJ

ΔH° = – 283.0 kJ

ΔH° = – 393.5 kJ

+ ½O2

C + ½O2 → CO ΔH° = – 110.5 kJCO + ½O2 → CO2 ΔH° = – 283.0 kJ

Practice Exercise 6 (pg. 167) with DiagramUsing example 5.6 as a model, try PE 6.

Draw the related enthalpy diagram.

5.51 (pg. 175)

5.52 (pg. 175)

Hess’s law: Example 5.7 (pg. 166)We may need to manipulate equations further:

2Fe + 1.5O2 → Fe2O3 ΔH°=?, givenFe2O3 + 3CO → 2Fe + 3CO2 ΔH°= – 26.74 kJ

CO + ½O2 → CO2 ΔH°= –282.96 kJ1: Align equations based on reactants/products.2: Multiply based on final reaction.3: Add equations.

2Fe + 1.5O2 → Fe2O3

3CO + 1.5O2 → 3CO2 ΔH°= –848.88 kJ

2Fe + 3CO2 → Fe2O3 + 3CO ΔH°= + 26.74 kJCO + ½O2 → CO2 ΔH°= –282.96 kJ

Don’t forget to add states. Try 5.55, 5.57, 5.58, 5.61 (pg. 175)ΔH°= –822.14 kJ